Question

Question #12d17

Answer 1

26) and 27)

Explanation:

26) We know that
`cos(x-y)=Cos(x) Cos(y) + Sin(x) Sin(y)`
and
`cos^2x+sin^2x=1`
then substituting
`cos(x-y) = (sqrt[1 - (3/5)^2]) (5/13) + (3/5) sqrt[1 - (5/13)^2]`

27) `cos(2x) = sin x = cos(pi/2-x)`
so
`2x=pi/2-x+2k pi`
solving for `x`
`x = pi/6+2/3k pi` for `k = 0,pm1,pm2,...`